Thermo 5th Chap06 P001

6-1

Chapter 6 THE SECOND LAW OF THERMODYNAMICS

The Second Law of Thermodynamics and Thermal Energy Reservoirs 6-1C Water is not a fuel; thus the claim is false. 6-2C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity. 6-3C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to a room. 6-4C Transferring 5 kWh of heat to an electric resistance wire and producing 6 kWh of electricity. 6-5C No. Heat cannot flow from a low-temperature medium to a higher temperature medium. 6-6C A thermal-energy reservoir is a body that can supply or absorb finite quantities of heat isothermally. Some examples are the oceans, the lakes, and the atmosphere. 6-7C Yes. Because the temperature of the oven remains constant no matter how much heat is transferred to the potatoes. 6-8C The surrounding air in the room that houses the TV set. Heat Engines and Thermal Efficiency 6-9C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics. 6-10C Heat engines are cyclic devices that receive heat from a source, convert some of it to work, and reject the rest to a sink. 6-11C Method (b). With the heating element in the water, heat losses to the surrounding air are minimized, and thus the desired heating can be achieved with less electrical energy input. 6-12C No. Because 100% of the work can be converted to heat. 6-13C It is expressed as "No heat engine can exchange heat with a single reservoir, and produce an equivalent amount of work". 6-14C (a) No, (b) Yes. According to the second law, no heat engine can have and efficiency of 100%. 6-15C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics. 6-16C No. The Kelvin-Plank limitation applies only to heat engines; engines that receive heat and convert some of it to work.

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

6-2

6-17 The power output and thermal efficiency of a power plant are given. The rate of heat rejection is to be determined, and the result is to be compared to the actual case in practice. Assumptions 1 The plant operates steadily. 2 Heat losses from the working fluid at the pipes and other components are negligible. Analysis The rate of heat supply to the power plant is determined from the thermal efficiency relation,

MW 15000.4

MW 600

th

outnet, ===η

WQH

&&

sink

Furnace

HE ηth = 40%The rate of heat transfer to the river water is determined from the first law

relation for a heat engine, 600 MW

MW 900=−=−= 6001500outnet,WQQ HL&&&

In reality the amount of heat rejected to the river will be lower since part of the heat will be lost to the surrounding air from the working fluid as it passes through the pipes and other components. 6-18 The rates of heat supply and heat rejection of a power plant are given. The power output and the thermal efficiency of this power plant are to be determined. Assumptions 1 The plant operates steadily. 2 Heat losses from the working fluid at the pipes and other components are taken into consideration. Analysis (a) The total heat rejected by this power plant is

LQ&

GJ/h 280=HQ&

HE

Furnace

sink

GJ/h 1538145 =+=LQ&

Then the net power output of the plant becomes

MW35.3 GJ/h 127153280outnet, ==−=−= LH QQW &&&

(b) The thermal efficiency of the plant is determined from its definition,

45.4%==== 0.454GJ/h 280GJ/h 127outnet,

thHQ

W&

&η


6-3

6-19E The power output and thermal efficiency of a car engine are given. The rate of fuel consumption is to be determined. Assumptions The car operates steadily. Properties The heating value of the fuel is given to be 19,000 Btu/lbm. Analysis This car engine is converting 28% of the chemical energy released during the combustion process into work. The amount of energy input required to produce a power output of 110 hp is determined from the definition of thermal efficiency to be

Btu/h ,598999hp 1Btu/h 2545

0.28hp 110outnet, =

==

thH

WQ

η

&&

To supply energy at this rate, the engine must burn fuel at a rate of

lbm/h 52.6==Btu/lbm 19,000

Btu/h 999,598m&

since 19,000 Btu of thermal energy is released for each lbm of fuel burned.

110 hp

28%

Engine Fuel

19,000 Btu/lbm HE

sink

6-20 The power output and fuel consumption rate of a power plant are given. The thermal efficiency is to be determined. Assumptions The plant operates steadily.

150 MW

Furnace 60 t/h

coal HE

sink

Properties The heating value of coal is given to be 30,000 kJ/kg. Analysis The rate of heat supply to this power plant is

( )( ) MW 500kJ/h 101.8kJ/kg 30,000kg/h 60,000 9coalcoal =×=== umQH &&

Then the thermal efficiency of the plant becomes

30.0%==== 0.300MW 500MW 150outnet,

thHQ

W&

&η

6-21 The power output and fuel consumption rate of a car engine are given. The thermal efficiency of the engine is to be determined. Assumptions The car operates steadily. Properties The heating value of the fuel is given to be 44,000 kJ/kg. Analysis The mass consumption rate of the fuel is

kg/h 22.4)L/h 28)(kg/L 0.8()( fuelfuel === Vm && ρ

The rate of heat supply to the car is

kW 273.78kJ/h 985,600)kJ/kg 44,000)(kg/h 22.4(coalcoal ==== umQH &&

60 kW

Engine Fuel

m =28 L/h HE

sink

Then the thermal efficiency of the car becomes

21.9%==== 0.219kW 273.78

kW 60outnet,th

HQW&

&η


6-4

6-22E The power output and thermal efficiency of a solar pond power plant are given. The rate of solar energy collection is to be determined.

350 kW

4%

Source

Solar pond HE

sink

Assumptions The plant operates steadily. Analysis The rate of solar energy collection or the rate of heat supply to the power plant is determined from the thermal efficiency relation to be

Btu/h102.986 7×=

==

h 1s 3600

kJ 1.055Btu 1

0.04kW 350

th

outnet,

ηW

QH

&&

6-23 The United States produces about 51 percent of its electricity from coal at a conversion efficiency of about 34 percent. The amount of heat rejected by the coal-fired power plants per year is to be determined. Analysis Noting that the conversion efficiency is 34%, the amount of heat rejected by the coal plants per year is

kWh 103.646 12×=×−×

=−=

+==

kWh 10878.134.0

kWh 10878.1

1212

coalth

coalout

coalout

coal

in

coalth

WW

Q

WQW

QW

η

η

outQ&

1.878x1012 kWh

ηth = 34%

Furnace Coal

HE

sink


6-5

6-24 The projected power needs of the United States is to be met by building inexpensive but inefficient coal plants or by building expensive but efficient IGCC plants. The price of coal that will enable the IGCC plants to recover their cost difference from fuel savings in 5 years is to be determined. Assumptions 1 Power is generated continuously by either plant at full capacity. 2 The time value of money (interest, inflation, etc.) is not considered. Properties The heating value of the coal is given to be 28×106 kJ/ton. Analysis For a power generation capacity of 150,000 MW, the construction costs of coal and IGCC plants and their difference are

Constructi 999

9IGCC

9coal

1030$10195$10225$differencecost on Constructi

10$225=kW)kW)($1500/ 000,000,150(coston

10$195=kW)kW)($1300/ 000,000,150(coston Constructi

×=×−×=

×=

×=

The amount of electricity produced by either plant in 5 years is

kWh 106.570=h) 24365kW)(5 000,000,150( 12×××=∆= tWWe&

The amount of fuel needed to generate a specified amount of power can be determined from

) valueHeating( valueHeating

or infuelin

in ηηη eee WQ

mW

QQW

===→=

Then the amount of coal needed to generate this much electricity by each plant and their difference are

tons10607.0=10877.110484.2

tons10877.1kWh 1

kJ 3600kJ/ton) 1028)(45.0(kWh 10570.6

) valueHeating(

tons10484.2kWh 1

kJ 3600kJ/ton) 1028)(34.0(kWh 10570.6

) valueHeating(

999plant IGCC coal,plant coal coal,coal

96

12

plant IGCC coal,

96

12

plant coal coal,

××−×=−=∆

×=

×

×==

×=

×

×==

mmm

Wm

Wm

e

e

η

η

For to pay for the construction cost difference of $30 billion, the price of coal should be ∆mcoal

$49.4/ton=×

×=

∆=

tons10607.01030$differencecost on Constructicoal ofcost Unit

9

9

coalm

Therefore, the IGCC plant becomes attractive when the price of coal is above $49.4 per ton.


6-6

6-25 EES Problem 6-24 is reconsidered. The price of coal is to be investigated for varying simple payback periods, plant construction costs, and operating efficiency. Analysis The problem is solved using EES, and the solution is given below. "Knowns:" HeatingValue = 28E+6 [kJ/ton] W_dot = 150E+6 [kW] {PayBackPeriod = 5 [years] eta_coal = 0.34 eta_IGCC = 0.45 CostPerkW_Coal = 1300 [$/kW] CostPerkW_IGCC=1500 [$/kW]} "Analysis:" "For a power generation capacity of 150,000 MW, the construction costs of coal and IGCC plants and their difference are" ConstructionCost_coal = W_dot *CostPerkW_Coal ConstructionCost_IGCC= W_dot *CostPerkW_IGCC ConstructionCost_diff = ConstructionCost_IGCC - ConstructionCost_coal "The amount of electricity produced by either plant in 5 years is " W_ele = W_dot*PayBackPeriod*convert(year,h) "The amount of fuel needed to generate a specified amount of power can be determined from the plant efficiency and the heating value of coal." "Then the amount of coal needed to generate this much electricity by each plant and their difference are" "Coal Plant:" eta_coal = W_ele/Q_in_coal Q_in_coal = m_fuel_CoalPlant*HeatingValue*convert(kJ,kWh) "IGCC Plant:" eta_IGCC = W_ele/Q_in_IGCC Q_in_IGCC = m_fuel_IGCCPlant*HeatingValue*convert(kJ,kWh) DELTAm_coal = m_fuel_CoalPlant - m_fuel_IGCCPlant "For to pay for the construction cost difference of $30 billion, the price of coal should be" UnitCost_coal = ConstructionCost_diff /DELTAm_coal "Therefore, the IGCC plant becomes attractive when the price of coal is above $49.4 per ton. " SOLUTION ConstructionCost_coal=1.950E+11 [dollars] ConstructionCost_diff=3.000E+10 [dollars] ConstructionCost_IGCC=2.250E+11 [dollars] CostPerkW_Coal=1300 [dollars/kW] CostPerkW_IGCC=1500 [dollars/kW] DELTAm_coal=6.073E+08 [tons] eta_coal=0.34 eta_IGCC=0.45 HeatingValue=2.800E+07 [kJ/ton] m_fuel_CoalPlant=2.484E+09 [tons] m_fuel_IGCCPlant=1.877E+09 [tons] PayBackPeriod=5 [years] Q_in_coal=1.932E+13 [kWh] Q_in_IGCC=1.460E+13 [kWh] UnitCost_coal=49.4 [dollars/ton] W_dot=1.500E+08 [kW] W_ele=6.570E+12 [kWh]


6-7

Following is a study on how unit cost of fuel changes with payback period:

PaybackPeriod [years]

UnitCostcoal [$/ton]

1 247 2 123.5 3 82.33 4 61.75 5 49.4 6 41.17 7 35.28 8 30.87 9 27.44 10 24.7

1 2 3 4 5 6 7 8 9 100

50

100

150

200

250

PayBackPeriod [years]

UnitCostcoal

[$/ton]

6-26 The projected power needs of the United States is to be met by building inexpensive but inefficient coal plants or by building expensive but efficient IGCC plants. The price of coal that will enable the IGCC plants to recover their cost difference from fuel savings in 3 years is to be determined. Assumptions 1 Power is generated continuously by either plant at full capacity. 2 The time value of money (interest, inflation, etc.) is not considered. Properties The heating value of the coal is given to be 28×106 kJ/ton. Analysis For a power generation capacity of 150,000 MW, the construction costs of coal and IGCC plants and their difference are

999

9IGCC

9coal

1030$10195$10225$differencecost on Constructi



×=×−×=

×=

×=

The amount of electricity produced by either plant in 3 years is kWh 103.942=h) 24365kW)(3 000,000,150( 12×××=∆= tWWe

&

The amount of fuel needed to generate a specified amount of power can be determined from

) valueHeating( valueHeating

or infuelin

in ηηη eee WQ

mW

QQW

===→=

Then the amount of coal needed to generate this much electricity by each plant and their difference are

tons10365.0=10126.110491.1

tons10126.1kWh 1

kJ 3600kJ/ton) 1028)(45.0(kWh 10942.3

) valueHeating(

tons10491.1kWh 1

kJ 3600kJ/ton) 1028)(34.0(kWh 10942.3

) valueHeating(

999plant IGCC coal,plant coal coal,coal

96

12

plant IGCC coal,

96

12

plant coal coal,

××−×=−=∆

×=

×

×==

×=

×

×==

mmm

Wm

Wm

e

e

η

η

For to pay for the construction cost difference of $30 billion, the price of coal should be ∆mcoal

$82.2/ton=×

×=

∆=

tons10365.01030$differencecost on Constructicoal ofcost Unit

9

9

coalm

Therefore, the IGCC plant becomes attractive when the price of coal is above $82.2 per ton.


6-8

6-27E An OTEC power plant operates between the temperature limits of 86°F and 41°F. The cooling water experiences a temperature rise of 6°F in the condenser. The amount of power that can be generated by this OTEC plans is to be determined. Assumptions 1 Steady operating conditions exist. 2 Water is an incompressible substance with constant properties. Properties The density and specific heat of water are taken ρ = 64.0 lbm/ft3 and C = 1.0 Btu/lbm.°F, respectively. Analysis The mass flow rate of the cooling water is

lbm/s 1897=lbm/min 113,790=gal 7.4804

ft 1gal/min) )(13,300lbm/ft 0.64(3

3waterwater

== Vm && ρ

The rate of heat rejection to the cooling water is

Btu/s 11,380=F)F)(6Btu/lbm. lbm/s)(1.0 1897()( inoutwaterout °°=−= TTCmQ &&

Noting that the thermal efficiency of this plant is 2.5%, the power generation is determined to be

kW 308=Btu/s 292 Btu/s) 380,11(

0.025 outin

=→+

=→+

== WW

WQW

WQW &

&

&

&&

&

&

&η

since 1 kW = 0.9478 Btu/s. 6-28 A coal-burning power plant produces 300 MW of power. The amount of coal consumed during a one-day period and the rate of air flowing through the furnace are to be determined. Assumptions 1 The power plant operates steadily. 2 The kinetic and potential energy changes are zero. Properties The heating value of the coal is given to be 28,000 kJ/kg. Analysis (a) The rate and the amount of heat inputs to the power plant are

MW 5.93732.0MW 300

th

outnet,in ===

ηW

Q&

&

MJ 101.8s) 360024(MJ/s) 5.937( 7inin ×=×=∆= tQQ &

The amount and rate of coal consumed during this period are

kg/s 48.33s 360024kg 10893.2

MJ/kg 28MW 101.8

6coal

coal

7

HV

incoal

=××

=∆

=

×=×

==

tm

m

qQ

m

&

kg 102.893 6

(b) Noting that the air-fuel ratio is 12, the rate of air flowing through the furnace is kg/s 401.8=== kg/s) 48.33(fuel) air/kg kg 12(AF)( coalair mm &&


6-9

Refrigerators and Heat Pumps 6-29C The difference between the two devices is one of purpose. The purpose of a refrigerator is to remove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a warm medium. 6-30C The difference between the two devices is one of purpose. The purpose of a refrigerator is to remove heat from a refrigerated space whereas the purpose of an air-conditioner is remove heat from a living space. 6-31C No. Because the refrigerator consumes work to accomplish this task. 6-32C No. Because the heat pump consumes work to accomplish this task. 6-33C The coefficient of performance of a refrigerator represents the amount of heat removed from the refrigerated space for each unit of work supplied. It can be greater than unity. 6-34C The coefficient of performance of a heat pump represents the amount of heat supplied to the heated space for each unit of work supplied. It can be greater than unity. 6-35C No. The heat pump captures energy from a cold medium and carries it to a warm medium. It does not create it. 6-36C No. The refrigerator captures energy from a cold medium and carries it to a warm medium. It does not create it. 6-37C No device can transfer heat from a cold medium to a warm medium without requiring a heat or work input from the surroundings. 6-38C The violation of one statement leads to the violation of the other one, as shown in Sec. 6-4, and thus we conclude that the two statements are equivalent. 6-39 The COP and the refrigeration rate of a refrigerator are given. The power consumption and the rate of heat rejection are to be determined. Assumptions The refrigerator operates steadily. Analysis (a) Using the definition of the coefficient of performance, the power input to the refrigerator is determined to be

LQ&

cool space

Kitchen air

R COP

kW 0.83==== kJ/min 051.2kJ/min 60

COPRinnet,

LQW&

&

(b) The heat transfer rate to the kitchen air is determined from the energy balance,

kJ/min 110=+=+= 5060innet,WQQ LH&&&


6-10

6-40 The power consumption and the cooling rate of an air conditioner are given. The COP and the rate of heat rejection are to be determined. Assumptions The air conditioner operates steadily. Analysis (a) The coefficient of performance of the air-conditioner (or refrigerator) is determined from its definition,

kJ/min 750=LQ&

House

Outdoors

A/C 6 kW

2.08=

==

kJ/min 60kW 1

kW 6kJ/min 750COP

innet,R W

QL&

&

(b) The rate of heat discharge to the outside air is determined from the energy balance,

Q ( ) kJ/min 1110=×+=+= kJ/min) 606(kJ/min 750innet,WQLH&&&

6-41 The COP and the refrigeration rate of a refrigerator are given. The power consumption of the refrigerator is to be determined. Assumptions The refrigerator operates steadily.

800 kJ/hCOP =

R

Kitchen air

Refrigerator

Analysis Since the refrigerator runs one-fourth of the time and removes heat from the food compartment at an average rate of 800 kJ/h, the refrigerator removes heat at a rate of

kJ/h 3200)kJ/h 800(4 =×=LQ&

when running. Thus the power the refrigerator draws when it is running is

kW 0.40==== kJ/h 14552.2

kJ/h 3200COPR

innet,LQW&

&

6-42E The COP and the refrigeration rate of an ice machine are given. The power consumption is to be determined. Assumptions The ice machine operates steadily.

&QL

ice 25°F

water55°F

COP = 2.4

R

Outdoors

Ice Machine

Analysis The cooling load of this ice machine is ( )( ) Btu/h 4732Btu/lbm 169lbm/h 28 === LL qmQ &&

Using the definition of the coefficient of performance, the power input to the ice machine system is determined to be

hp 0.775=

==

Btu/h 2545hp 1

2.4Btu/h 4732

COPRinnet,

LQW&

&


6-11

6-43 The COP and the power consumption of a refrigerator are given. The time it will take to cool 5 watermelons is to be determined. Assumptions 1 The refrigerator operates steadily. 2 The heat gain of the refrigerator through its walls, door, etc. is negligible. 3 The watermelons are the only items in the refrigerator to be cooled. Properties The specific heat of watermelons is given to be c = 4.2 kJ/kg.°C. Analysis The total amount of heat that needs to be removed from the watermelons is

( ) ( )( )( ) kJ 2520C820CkJ/kg 4.2kg 105swatermelon =−°⋅×=∆= oTmcQL

The rate at which this refrigerator removes heat is

450 WCOP = 2.5

R

Kitchen air

cool space

( )( ) ( )( ) kW 1.125kW 0.452.5COP innet,R === WQL&&

That is, this refrigerator can remove 1.125 kJ of heat per second. Thus the time required to remove 2520 kJ of heat is

min37.3 s 2240kJ/s 1.125kJ 2520

====∆L

L

QQt &

This answer is optimistic since the refrigerated space will gain some heat during this process from the surrounding air, which will increase the work load. Thus, in reality, it will take longer to cool the watermelons. 6-44 [Also solved by EES on enclosed CD] An air conditioner with a known COP cools a house to desired temperature in 15 min. The power consumption of the air conditioner is to be determined. Assumptions 1 The air conditioner operates steadily. 2 The house is well-sealed so that no air leaks in or out during cooling. 3 Air is an ideal gas with constant specific heats at room temperature. Properties The constant volume specific heat of air is given to be cv = 0.72 kJ/kg.°C. Analysis Since the house is well-sealed (constant volume), the total amount of heat that needs to be removed from the house is ( ) ( )( )( ) kJ 6912C2032CkJ/kg 0.72kg 800House =°−°⋅=∆= TmcQL v

HQ&

32→20°C

House

COP = 2.5

Outside

AC

This heat is removed in 15 minutes. Thus the average rate of heat removal from the house is

kW 7.68s 6015

kJ 6912=

×=

∆=

tQQ L

L&

Using the definition of the coefficient of performance, the power input to the air-conditioner is determined to be

kW3.07 2.5

kW 7.68COPR

innet, === LQW&

&


6-12

6-45 EES Problem 6-44 is reconsidered. The rate of power drawn by the air conditioner required to cool the house as a function for air conditioner EER ratings in the range 9 to 16 is to be investigated. Representative costs of air conditioning units in the EER rating range are to be included. Analysis The problem is solved using EES, and the results are tabulated and plotted below. "Input Data" T_1=32 [C] T_2=20 [C] C_v = 0.72 [kJ/kg-C] m_house=800 [kg] DELTAtime=20 [min] {SEER=9} COP=EER/3.412 "Assuming no work done on the house and no heat energy added to the house in the time period with no change in KE and PE, the first law applied to the house is:" E_dot_in - E_dot_out = DELTAE_dot E_dot_in = 0 E_dot_out = Q_dot_L DELTAE_dot = m_house*DELTAu_house/DELTAtime DELTAu_house = C_v*(T_2-T_1) "Using the definition of the coefficient of performance of the A/C:" W_dot_in = Q_dot_L/COP "kJ/min"*convert('kJ/min','kW') "kW" Q_dot_H= W_dot_in*convert('KW','kJ/min') + Q_dot_L "kJ/min"

EER [Btu/kWh]

Win [kW]

9 2.184 10 1.965 11 1.787 12 1.638 13 1.512 14 1.404 15 1.31 16 1.228

9 10 11 12 13 14 15 161.2

1.4

1.6

1.8

2

2.2

EER [Btu/kWh]

Win

[kW]


6-13

6-46 The heat removal rate of a refrigerator per kW of power it consumes is given. The COP and the rate of heat rejection are to be determined. Assumptions The refrigerator operates steadily. Analysis The coefficient of performance of the refrigerator is determined from its definition,

1.4=

==

kJ/h 3600kW 1

kW 1kJ/h 5040COP

innet,R W

QL&

&

The rate of heat rejection to the surrounding air, per kW of power consumed, is determined from the energy balance,

( ) kJ/h 8640=×+=+= kJ/h 3600(1kJ/h) 5040innet,WQQ LH&&&

5040 kJ/h1 kW

R

Kitchen air

Refrigerator

6-47 The rate of heat supply of a heat pump per kW of power it consumes is given. The COP and the rate of heat absorption from the cold environment are to be determined. Assumptions The heat pump operates steadily.

8000 kJ/h

1 kWHP

House

Outside

Analysis The coefficient of performance of the refrigerator is determined from its definition,

2.22=

==

kJ/h 3600kW 1

kW 1kJ/h 8000COP

innet,HP W

QH&

&

The rate of heat absorption from the surrounding air, per kW of power consumed, is determined from the energy balance,

( ) kJ/h 4400=−=−= kJ/h (1)(3600kJ/h) 8,000innet,WQQ HL&&&

6-48 A house is heated by resistance heaters, and the amount of electricity consumed during a winter month is given. The amount of money that would be saved if this house were heated by a heat pump with a known COP is to be determined. Assumptions The heat pump operates steadily. Analysis The amount of heat the resistance heaters supply to the house is equal to he amount of electricity they consume. Therefore, to achieve the same heating effect, the house must be supplied with 1200 kWh of energy. A heat pump that supplied this much heat will consume electrical power in the amount of

kWh 5002.4

kWh 1200COPHP

innet, === HQW&

&

which represent a savings of 1200 – 500 = 700 kWh. Thus the homeowner would have saved (700 kWh)(0.085 $/kWh) = $59.50


6-14

6-49E The rate of heat supply and the COP of a heat pump are given. The power consumption and the rate of heat absorption from the outside air are to be determined. Assumptions The heat pump operates steadily.

HQ&

60,000 Btu/h

COP = 2.5HP

House

Outside

Analysis (a) The power consumed by this heat pump can be determined from the definition of the coefficient of performance of a heat pump to be

hp9.43 Btu/h 24,0002.5

Btu/h 60,000COPHP

innet, ==== HQW&

&

(b) The rate of heat transfer from the outdoor air is determined from the conservation of energy principle,

( ) Btu/h 36,000=−=−= Btu/h24,00060,000innet,WQQ HL&&&

6-50 The rate of heat loss from a house and the COP of the heat pump are given. The power consumption of the heat pump when it is running is to be determined. Assumptions The heat pump operates one-third of the time. Analysis Since the heat pump runs one-third of the time and must supply heat to the house at an average rate of 22,000 kJ/h, the heat pump supplies heat at a rate of

Outside

House

HQ&

22,000kJ/h

COP = 2.8HP hkJ 000,66kJ/h) 22,000(3 /=×=HQ&

when running. Thus the power the heat pump draws when it is running is

kW 6.55=

==

kJ/h 3600kW 1

2.8kJ/h 66,000

COPHPinnet,

HQW&

&

6-51 The rate of heat loss, the rate of internal heat gain, and the COP of a heat pump are given. The power input to the heat pump is to be determined. Assumptions The heat pump operates steadily.

HQ&

60,000kJ/h

COP = 2.5HP

House

Outside

Analysis The heating load of this heat pump system is the difference between the heat lost to the outdoors and the heat generated in the house from the people, lights, and appliances,

& , ,QH = − =60 000 4 000 56 000 kJ h, /

Using the definition of COP, the power input to the heat pump is determined to be

kW6.22 kJ/h 3600

kW 12.5

kJ/h 56,000COPHP

innet, =

== HQW

&&


6-15

6-52E An office that is being cooled adequately by a 12,000 Btu/h window air-conditioner is converted to a computer room. The number of additional air-conditioners that need to be installed is to be determined. Assumptions 1 The computers are operated by 4 adult men. 2 The computers consume 40 percent of their rated power at any given time. Properties The average rate of heat generation from a person seated in a room/office is 100 W (given). Analysis The amount of heat dissipated by the computers is equal to the amount of electrical energy they consume. Therefore,

4000 Btu/h

Computer room

Outside

AC

Q Q

& (& ( ) & (& & &

Q

Q Q Q

computers

people person

total computers people

Rated power) (Usage factor) = (3.5 kW)(0.4) = 1.4 kW

No. of people W) W

W = 6142 Btu / h

= ×

= × = × =

= + = + =

4 100 400

1400 400 1800

since 1 W = 3.412 Btu/h. Then noting that each available air conditioner provides 4,000 Btu/h cooling, the number of air-conditioners needed becomes

rsconditione Air 2≈=

==

5.1Btu/h 4000Btu/h 6142

A/C ofcapacity Coolingload Coolingrsconditioneair of No.

6-53 A decision is to be made between a cheaper but inefficient air-conditioner and an expensive but efficient air-conditioner for a building. The better buy is to be determined. Assumptions The two air conditioners are comparable in all aspects other than the initial cost and the efficiency. Analysis The unit that will cost less during its lifetime is a better buy. The total cost of a system during its lifetime (the initial, operation, maintenance, etc.) can be determined by performing a life cycle cost analysis. A simpler alternative is to determine the simple payback period. The energy and cost savings of the more efficient air conditioner in this case is

kWh/year 13,500=)0.5/11/3.2kWh/year)( 000,120(

)COP/1COP/1)(load cooling Annual(B) of usageenergy Annual(A) of usageenergy Annual(savingsEnergy

BA

−=−=

−=

Air Cond. A COP = 3.2

$1350/year=$0.10/kWh)kWh/year)( 500,13(

energy) ofcost Unit )(savingsEnergy (savingsCost ==

The installation cost difference between the two air-conditioners is Cost difference = Cost of B – cost of A = 7000 – 5500 = $1500 Air Cond. B

COP = 5.0 Therefore, the more efficient air-conditioner B will pay for the $1500 cost differential in this case in about 1 year. Discussion A cost conscious consumer will have no difficulty in deciding that the more expensive but more efficient air-conditioner B is clearly the better buy in this case since air conditioners last at least 15 years. But the decision would not be so easy if the unit cost of electricity at that location was much less than $0.10/kWh, or if the annual air-conditioning load of the house was much less than 120,000 kWh.


6-16

6-54 Refrigerant-134a flows through the condenser of a residential heat pump unit. For a given compressor power consumption the COP of the heat pump and the rate of heat absorbed from the outside air are to be determined. Assumptions 1 The heat pump operates steadily. 2 The kinetic and potential energy changes are zero. Properties The enthalpies of R-134a at the condenser inlet and exit are

kJ/kg 47.95

0kPa 800

kJ/kg 22.271C35kPa 800

22

2

11

1

=

==

=

°==

hxP

hTP

Analysis (a) An energy balance on the condenser gives the heat rejected in the condenser

kW 164.3kJ/kg )47.9522.271(kg/s) 018.0()( 21 =−=−= hhmQH &&

Win

QL

800 kPa 35°C

Expansion valve Compressor

Evaporator

Condenser

800 kPax=0

QH

The COP of the heat pump is

2.64===kW 2.1

kW 164.3COPinW

QH&

&

(b) The rate of heat absorbed from the outside air

kW 1.96=−=−= 2.1164.3inWQQ HL&&&

6-55 A commercial refrigerator with R-134a as the working fluid is considered. The evaporator inlet and exit states are specified. The mass flow rate of the refrigerant and the rate of heat rejected are to be determined.

Win

QL

120 kPax=0.2

Expansion valve Compressor

Evaporator

Condenser

120 kPa-20°C

QHAssumptions 1 The refrigerator operates steadily. 2 The kinetic and potential energy changes are zero. Properties The properties of R-134a at the evaporator inlet and exit states are (Tables A-11 through A-13)

kJ/kg 84.238C20

kPa 120

kJ/kg 38.652.0

kPa 120

22

2

11

1

=

°−==

=

==

hTP

hxP

Analysis (a) The refrigeration load is

kW 54.0kW) 45.0)(2.1(COP)( in === WQL&&

The mass flow rate of the refrigerant is determined from

kg/s 0.0031=−

=−

=kJ/kg )38.6584.238(

kW 54.0

12 hhQ

m LR

&&

(b) The rate of heat rejected from the refrigerator is

kW 0.99=+=+= 45.054.0inWQQ LH&&&


6-17

Perpetual-Motion Machines 6-56C This device creates energy, and thus it is a PMM1. 6-57C This device creates energy, and thus it is a PMM1. Reversible and Irreversible Processes 6-58C No. Because it involves heat transfer through a finite temperature difference. 6-59C Because reversible processes can be approached in reality, and they form the limiting cases. Work producing devices that operate on reversible processes deliver the most work, and work consuming devices that operate on reversible processes consume the least work. 6-60C When the compression process is non-quasiequilibrium, the molecules before the piston face cannot escape fast enough, forming a high pressure region in front of the piston. It takes more work to move the piston against this high pressure region. 6-61C When an expansion process is non-quasiequilibrium, the molecules before the piston face cannot follow the piston fast enough, forming a low pressure region behind the piston. The lower pressure that pushes the piston produces less work. 6-62C The irreversibilities that occur within the system boundaries are internal irreversibilities; those which occur outside the system boundaries are external irreversibilities. 6-63C A reversible expansion or compression process cannot involve unrestrained expansion or sudden compression, and thus it is quasi-equilibrium. A quasi-equilibrium expansion or compression process, on the other hand, may involve external irreversibilities (such as heat transfer through a finite temperature difference), and thus is not necessarily reversible. The Carnot Cycle and Carnot's Principle 6-64C The four processes that make up the Carnot cycle are isothermal expansion, reversible adiabatic expansion, isothermal compression, and reversible adiabatic compression. 6-65C They are (1) the thermal efficiency of an irreversible heat engine is lower than the efficiency of a reversible heat engine operating between the same two reservoirs, and (2) the thermal efficiency of all the reversible heat engines operating between the same two reservoirs are equal. 6-66C False. The second Carnot principle states that no heat engine cycle can have a higher thermal efficiency than the Carnot cycle operating between the same temperature limits. 6-67C Yes. The second Carnot principle states that all reversible heat engine cycles operating between the same temperature limits have the same thermal efficiency. 6-68C (a) No, (b) No. They would violate the Carnot principle.


Thermo 5th Chap06 P001

Documents

Transcript of Thermo 5th Chap06 P001